A body of mass M text{ kg} is at the top poin | vedclass

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(C) Let the body leave the surface of the hemisphere at point $P$. At point $P$, let the radius vector of the body make an angle $	heta$ with the ver

Vedclass · Accepted Answer

$60$

Vedclass · Answer

(C) Let the body leave the surface of the hemisphere at point $P$. At point $P$, let the radius vector of the body make an angle $	heta$ with the vertical. The forces acting on the body at point $P$ are the gravitational force $mg$ (downwards) and the normal reaction $R$ (outwards). The component of gravitational force towards the center is $mg \cos 	heta$. The net centripetal force is provided by the difference between the radial component of gravity and the normal reaction: $mg \cos 	heta - R = \frac{mv^2}{r}$ When the body leaves the surface, the normal reaction $R$ becomes zero. Therefore, $mg \cos 	heta = \frac{mv^2}{r}$ Substituting the given values $v = 5 	ext{ m/s}$, $r = 5 	ext{ m}$, and $g = 10 	ext{ m/s}^2$: $\cos 	heta = \frac{v^2}{rg} = \frac{5^2}{5 	imes 10} = \frac{25}{50} = \frac{1}{2}$ Since $\cos 	heta = \frac{1}{2}$, we have $	heta = 60^{\circ}$.

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